Analyze & Evaluate a Truss

8/8/2019 Analyze & Evaluate a Truss

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#3

Analyze and Evaluate a Truss

Learning Activity #3:

Overview of the Activity

In this learning activity, we will analyze and evaluate one of the main trusses from the Grant Road Bridge.We will create a mathematical model of the truss, then use this model as the basis for a structural analysisaseries of mathematical calculations to determine the internal force in every member of the truss. We will alsouse the experimental data from Learning Activity #2 to determine the strength of each truss member. Finally we

will perform a structural evaluationa comparison of the internal forces and strengths, to determine whetheror not the truss can safely carry its prescribed loads.

Why?

Engineering design is an iterative process. To create an optimal design, the engineer must develop manydifferent alternative solutions, evaluate each one, and then select the alternative that best satises the designrequirements. But how are these alternative solutions evaluated? Engineers use many different criteria toevaluate a design; but in structural design, the most important of these criteria is the structures ability to carryload safely. In most cases, an evaluation of structural safety can only be done mathematically. It would beimpractical, uneconomical, and unsafe for the structural engineer to evaluate a bridge design by building afull-size prototype, then running heavy trucks across the structure to determine if it is strong enough. When astructure is built, it must be strong enough to carry its prescribed loads. The engineer must get it right the rsttime. For this reason, the structural engineer must be able to mathematically model, analyze, and evaluate thestructure with a high degree of accuracyand without the benet of prototype testing. In this activity, you willlearn how an engineer performs a structural evaluation. In Learning Activity #5, you will apply this process todesign your own truss bridge.


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Learning Objectives

As a result of this learning activity, you will be able to do the following:

n Calculate the components of a force vector.

n Add two force vectors together.

n Explain the following structural engineering concepts: free body diagram, equilibrium, structural model, sym-metry, static determinacy, stability, and factor of safety.

n Use the Method of Joints to calculate the internal force in every member in a truss.

n Determine the strength of every member in a truss.

n Evaluate a truss, to determine if it can carry a given load safely.

Key Terms

To successfully complete this learning activity, you must understand the following key terms and concepts fromLearning Activities #1 and #2:

truss deck internal force tensile strength

member load tension compressive strength

joint reaction compression failure

If you have forgotten any of these terms, it would be a good idea to review their denitions in the Glossary(Appendix D) before proceeding.

Information

Analysis

An analysis is an examination of a complex system, usually conducted by breaking the system down into itscomponent parts. Once they are identied, the component parts and their relationships to the system as a wholecan be studied in detail. For example, suppose your baseball team has been losing a lot of games, and you want togure out why. Your team is a complex system. There are a lot of possible reasons why it might not befunctioning as well as it could. To analyze the performance of the team, youll need to break itdown into its component parts. The obvious way to do this is to look at the teams individualmembersnine players and a coach. But the team can also be broken down by its functionshitting, pitching, elding, and base running. To perform the analysis, you would look at eachteam member and each function in detail. You would examine batting, pitching, and eldingstatistics, to determine whether poor performance in any of these areas might be responsiblefor the teams losing record. You might discover, for example, that the teams batting averageagainst left-handed pitching has been particularly poor. This important analysis result mightbe used as the basis for designing a practice regimen to correct the problem.

Structural Analysis

A structural analysis is a mathematical examination of a structure, conducted by breaking the structure downinto its component parts, then studying how each part performs and how each part contributes to the performanceof the structure as a whole. Usually, the products of a structural analysis are (1) reactions, (2) internal memberforces, and (3) deectionshow much the structure bends or sways when it is loaded. Like the analysis of yourbaseball team, structural analysis is often used to determine if the system is performing as intended and, if it is not,to correct the problem.


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There is an important difference between structural analysis and structural design. Structural analysis isconcerned with examining existing structures to determine if they can carry load safely. Structural design isconcerned with creating new structures to meet the needs of society. Though analysis and design are fun-damentally different activities, they are closely interrelatedanalysis is an integral part of the design process.Well see how analysis and design t together in Learning Activity #4.

To perform a structural analysis, we will apply a variety of mathematical tools from geometry, trigonometry,and algebra, as well as some basic concepts from physics. These concepts are reviewed in the followingsections.

Some Basic Concepts from Trigonometry

A truss is a structure composed of members arranged in interconnected triangles. For this reason, thegeometry of triangles is very important in structural analysis. To analyze a truss, we must be able to mathemati-cally relate the angles of a triangle to the lengths of its sides. These relationships are part of a branch of math-ematics called trigonometry. Here we will review some basic concepts from trigonometry that are essentialtools for truss analysis.

This diagram shows a right trianglea triangle with one of its three angles measuringexactly 90o. Sides a and b form the 90o angle. The other two angles, identied as

1and

2, are

always less than 90o. Side c, the side opposite the 90o angle, is always the longest of the three

sides. It is called the hypotenuse of the right triangle.

Thanks to an ancient Greek mathematician named Pythagoras, we can easily calculate thelength of the hypotenuse of a right triangle. The Pythagorean Theorem tells us that

The Pythagorean Theorem shows how the lengths of the sides of a right triangle are related to each other.But how are the lengths of the sides related to the angles? Consider the denitions of two key terms fromtrigonometrysine and cosine. Both denitions are based on the geometry of a right triangle, as shown above.The sineof an angle (abbreviated sin) is dened as the length of the opposite side divided by the length of

the hypotenuse. For example, the sine of the angle 1 would be calculated as

In this case, side a is designated as the opposite side, because it is farthest from the angle 1. For the angle

2

, the opposite sideisb; thus, the sine of 2

is

The cosineof an angle (abbreviated cos) is dened as the length of the adjacent side divided by thelength of the hypotenuse. Applying this denition to our example, we have


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F = 20N

x

y

= 50 o

F = 20N

x

y

x

y

= 50 o

Fy

F

Fx

x

y

Fy

F

Fx

x

y

F

Fx

x

y

It is important to recognize that the sine and cosine of an angle do not depend on theoverall size of the triangleonly on the relative lengths of its sides. In the diagram at right,three different right triangles (ABC, ADE, and AFG) are drawn with a common angle

1. It doesnt matter which of the three triangles you use to calculate the

sine and cosine of 1. Youll get the same answers in all three cases,

because the relative lengths of the sides are all the same.

Well see important applications of the sine and cosine when we analyze a truss, later in this learning activity.

Working with Vectors

A force can be represented as a vectora mathematical quantity that has both magnitude and direction. Whenwe perform a structural analysis, we will calculate both the magnitude and direction of every force that acts on thestructure. Thus, before when can analyze a structure, we need to learn how to work with vectors. Specically, weneed to learn two basic concepts from vector mathbreaking a vector into its components and adding vectorstogether.

Breaking a Vector into its Components

When we analyze a truss, we will need to describe the directions of force vectorsmathematically. To do this, we must rst dene a coordinate axis system. For a two-dimensional structure, we normally use an x-axis to represent the horizontal directionand a y-axis to represent the vertical. Once the coordinate axis system is established,we can represent thedirection of any vector as an angle measured from either thex-axis or the y-axis. For example, the force vector at right has a magnitude (F) of 20newtons and a direction () of 50 degrees, measured counterclockwise from thex-axis.

This force can also be represented as two equivalent forces, one in the x-directionand one in the y-direction. Each of these forces is called a component of the vector F.

To determine the magnitudes of these two components, visualize a right triangle withthe vector Fas the hypotenuse and the other two sides parallel to the x-axis and y-axis.If Fis the length of the hypotenuse, then the lengths of the two perpendicular sidesare exactly equal to the x-component and y-component of F. We use the symbol Fxtorepresent the x-component of Fand the symbol Fyto represent the y-component.


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From trigonometry, we can apply the denitions of the sine and the cosine to calculate the two components.Recall that

From the diagram on the previous page, we can see that Fyis the opposite side of the triangle, and Fis thehypotenuse. Substituting, we get

If we multiply both sides of this equation by F, we get

Similarly,

Therefore, if we know the magnitude (F) and direction () of a force, then we can use the equations above tocalculate the two components of the force.

The diagram at right shows the correct way to represent the force Fand itscomponentswith all three vectors originating from the same point. The twodotted lines show that Fxand Fyare the same lengths as the sides of a right trianglewith Fas its hypotenuse.

Returning to our example, if we substitute the actual numerical values F=20N and=50, and use a calculator to determine the sine and cosine of the angle, we get thefollowing results

The small arrows to the right of the answers indicate the directions of the Fyand Fxvectors. When we writea vector quantity, we must always be careful to show both its magnitude and direction.

But what do these numbers really mean? Suppose you kick asoccer ball with a single 20-newton force at an angle of 50o. Thisforce will cause the ball to move a particular direction and distance.Now suppose that two players kick the ball simultaneouslyonewith a 15.3-newton force in the y-direction and one with a12.9-newton force in the x-direction. In this case, the ball willrespond exactly as it did when you kicked it with the single 20-new-ton force. The ball will move the same direction and distance,because it feels exactly the same force. The two components of aforce are exactly equivalent to that force and will produce exactlythe same effect on an object.

The two components of a force are exactly

equivalent to that force.


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Adding Vectors TogetherWhen two or more forces are applied to an object, it is often necessary to calculate the total force on the object.

We calculate the total force by simply adding all of the individual force vectors together. To add vectors, however,we must follow an important rule:or

To add vectors whose directions are not the same, we must do the following:

n Break each vector into its equivalent x-component and y-component.

n Add all of the x-components together.

n Add all of the y-components together.

As an example, lets add the two forces F1and F2, shown at right. We

begin by calculating the components of the two vectors:

Again the direction of each vector component is indicated with anarrow. We must pay careful attention to these directions when we addcomponents together. Note that F1xand F2xpoint in opposite directions.The directions of F1yand F2yare also opposite.

When we add the x-components, we will assume that the directionindicated by the x-axis is positive. Then the sum of the two x-components is

In this equation,F

1xis positive, because it points to the rightthe same direction as the positive x-axis.F

2xisnegative, because it points to the leftopposite the direction of the positive x-axis. The answer is negative, whichmeans that the x-component of the total force is to the left. We write the nal answer as

The magnitudes of two or more vectors can be added together only if their directions are the same.


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* Actually, there are three equilibrium conditions for a two-dimensional structure. In addition to the two described above,

the sum of the moments about any point must also equal zero. The concept of a moment is a very important one; however,

it is beyond the scope of this book. The problems used in this and subsequent learning activities have been chosen so that

this third equilibrium condition is not required to obtain a correct solution.

Assuming that the direction of the positive y-axis (upward) is positive, the sum of the y-components is

In this case, the total is positive, so we conclude that the y-component of the total force is upward.

The total force and its two components are illustrated at right. If weneeded to know the actual magnitude of FTOTALwe could calculate it by

using the Pythagorean Theorem; however, for this learning activity, wewill only need to calculate the total x-component and the total y-com-ponent, as shown here.

Equilibrium

In Learning Activity #1, we dened equilibrium as a condition in which the total force acting on an object iszero. Now that we know how to actually calculate the total force on an object, we can apply the concept ofequilibrium as a powerful problem-solving tool. Specically, if we know that an object is in equilibriumbecause it is not movingthen we know that the total force on that object is zero; and we can use this fact tocalculate the magnitude and direction of unknown forces acting on the object.

Because we calculate total force by adding up the x-components and y-components separately, there are reallytwo conditions that must be satised if an object is in equilibrium.*

First, the sum of the x-components of all forces acting on the structure must be zero.We write this condition as

where the symbol means the sum of, and the entire expression is read, The sum of the forces in thex-direction equals zero.

The second equilibrium condition is that the sum of all forces in the y-direction must equal zero,which we write as

These two equations are commonly known as the equations of equilibrium. They are simple yet powerfulmathematical tools, with many different applications in science and engineering. In this learning activity, theequations of equilibrium will enable us to calculate the reactions and internal member forces in a truss.

Creating a Structural Model

A structural model is a mathematical idealization of a structurea series of simplifying assumptions aboutthe structures conguration and loading that allow us to predict its behavior mathematically.

When we model a two-dimensional truss, we typically make the following general assumptions:

n The truss members are perfectly straight.

n The joints that connect the truss members together are frictionless pins.

n Loads and reactions are applied only at the joints.


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Free Body Diagram of the

nutcracker truss

Taken together, these assumptions imply that the members of a truss do not bend. Truss members are assumedto carry load either in pure tension or in pure compression. These assumptions allow us to use a simple type ofstructural analysis that ignores the effects of bending.

None of these assumptions is perfectly accurate, however. In an actual truss bridge, members are never per-fectly straight, due to minor variations in the manufacturing and fabrication processes. Modern trusses use gussetplate connections, which do not behave like pins; and even in older bridges with pinned connections, the pins arecertainly not frictionless. Furthermore, actual trusses can never be loaded entirely at the joints, if only because theweight of the members themselves is distributed throughout the structure. Fortunately, the inaccuracies in our

assumptions generally produce only minor inaccuracies in our structural analysis results, and experienced engi-neers know how to compensate for these small errors to ensure the safety of their designs.

Having made these general assumptions about the structure, we must also make a number of specic decisionsabout how to represent the particular truss we are modeling. These decisions include:

n The geometric conguration of the truss, including the locations of all joints, the conguration ofthe members, and all relevant dimensions.

n The conguration of the supports.

n The magnitude and direction of the loads that will be applied to the structure.

Once we have decided how we will represent the structure, supports, and loads, we should always completethe modeling process by creating one or more drawings that clearly illustrate the structural model.

The Free Body Diagram

One of the most important tools in structural engineering is a simple sketch called the free body diagram.A free body diagram is a drawing of a bodya structure or a portion of a structureshowing all of the forcesacting on it. Drawing a free body diagram is the essential rst step in any structural analysis.

To draw a free body diagram:

n Draw the outline of the structure, completely isolated from its surroundings. Do not show any of the supportsthat connect the structure to its foundations.

n At the location of each support, draw and label the appropriate reactions.

n Draw and label all of the loads applied to the structure.

n Draw all relevant dimensions.

n Draw the x-y coordinate axis system.

In Learning Activity #1, we built a simple three-member truss by tying a shortpiece of string to the handles of a nutcracker. We then applied a 10-newton down-ward load to the top of the structure. The free body diagram for our nutcracker trussis shown here. Note that the downward 10-newton load is resisted by two upwardreactions at the bottom ends of the handles. Because the magnitudes of these forcesare unknown, they are labeled RA and RB. The three joints of the truss are labeled A,B, and C, for future reference.


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Calculating Reactions

Reactions are forces developed at the supports of a structure, to keep the structure in equilibrium. Giventhat the reactionsRA and RBon our nutcracker truss are in the y-direction, we can determine their magnitudeusing the equilibrium equation

Assuming that the upward direction is positive, the sum of the forces in the y-direction is

or

Because the structure, the load, and the supports are all symmetrical about the centerline of the nutcracker,the two reactions labeled RA and RBmust be equal. To understand why this is true, try the simple experimentillustrated below. Set up two scales a few inches apart, and lay a book across them. Ensure that the book iscentered between the two scales. Then place a relatively heavy object like a full can of soup on top of the book.Gradually slide the soup can from one end of the book to the other, and watch the readings on the two scales asyou move the can. You will notice that, when the can is perfectly centered between the two scales, the readingson the scales are exactly equal. At any other position, the readings are unequal. In this experiment, the soupcan is the load; the book is the structure, and the scales directly measure the two reactions. The experimentclearly demonstrates that the reactions are equal if the loads, the supports, and the structure itself are sym-metrical about a vertical centerline.

SOUP

SCALE SCALE

SOUP

SCALE SCALE

SOUP

SCALE SCALE

SOUPSOUP

SCALE SCALE

Centerline

Centerline

Centerline

Centerline

When the load is centered, the readings on the two scales are equal. Otherwise, the readings are unequal.


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along the centerlines of their respective members. Remember that we have only assumedFABand FACto be intension. We will check this assumption when we solve the equations of equilibrium in Step 3.

Step 3: Write and solve the equations of equilibrium.When a structure is in equilibrium, every part of that structure must also be in equilibrium. We know that

our nutcracker truss is in equilibrium, because it isnt moving; therefore, Joint A must be in equilibrium as well.Because Joint A is in equilibrium, we can write its two equilibrium equations. Lets start with the sum of forcesacting in the y-direction. To write this equation, look at the free body diagram of Joint A, and identify every

force that acts in the y-direction or has a component in the y-direction. Each of these forces must appear in theequilibrium equation. Assuming that the upward direction is positive,

To write this equation, it was necessary to break the force vector FACinto itsx-component and y-component, as shown at right. The y-component is FACsin65, andbecause it points upward, it is positive. The x-component is FACcos65, but this com-ponent does not appear in the Fyequilibrium equation, because it does not act inthe y-direction.

Since this equation has only one unknown variable, we can calculate sin65 andsolve for FACdirectly:

Because the answer is negative, our initial assumption about the direction of FACmust have been incorrect.

We assumed that the force FACis in tension; the negative answer tells us it is in compression. We can now writethe nal answer as

Note that, for internal forces, we do not show the direction of the force vector with an arrow; rather wesimply label the force as either tension or compression.

Now we can write the second equilibrium equationthe sum of the forces in the x-directionfor Joint A.Again look at the free body diagram of the joint, and identify every force that acts in the x-direction or has acomponent in the x-direction. Include each of these forces in the equilibrium equation:

Note that the x-component of the force FACappears in this equation, while the y-component does not.To solve the equation, we must calculate cos65, substitute the value of FACwe calculated above, and thensolve for FAB.


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A statically indeterminate truss.

A statically determinate truss.

An unstable truss.

The loads applied to structures are highly unpredictable.

wherejis the number of jointsand mis the number of members. For example,our nutcracker truss has 3 members and 3 joints. Substituting these numbersinto the equation above, we nd that 2jand m+3are both equal to6, so the

mathematical condition for static determinacy issatised. If 2j is less than m+3,then the truss is statically indeterminate. For example, the truss at right has 6joints and 10 members. Thus 2jis 12, and m+3is 13. Since 2jis less than m+3, thestructure is indeterminate. Such a structure cannot be analyzed using theequations of equilibrium alone. If you tried to use the Method of Joints to

analyze this truss, you would nd that you have more unknown forces than youhave equations available to solve for them. It is possible to analyze a staticallyindeterminate structure, but the solution process requires advanced engineer-ing concepts that are beyond the scope of this book.

If 2j is greater than m+3, then the truss is unstable. An unstable truss does nothave enough members to form a rigid framework. Such a structure cannot carryany load.

In general, a truss is stable if all of its members are arranged in a network ofinterconnected triangles. For example, the simple truss at right is composed of 6joints and 9 members, which together form four interconnected triangles (ABF,BCF, CEF, and CDE). This truss also satises the mathematical condition for

determinacy, since both 2jand m+3are equal to 12.

If member CF is removed, however, the truss becomes unstable. Without itsdiagonal member, the center panel of the truss now consists of a rectangle(BCEF) formed by four members, rather than two triangles (BCF and CEF). Thisconguration is unstable because there is nothing to prevent the rectangle BCEFfrom distorting into a parallelogram, as shown below. (Remember that weassume all truss joints to be frictionless pins.) For this truss, 2jis still 12, whilem+3is only 11. Since2j is greater than m+3, the mathematical test conrms ourobservation that the truss is unstable.

As you might expect from this example, an unstable truss can generally bemade stable by simply adding members until an appropriate arrangement of

interconnected triangles is achieved.

Factor of Safety

When an engineer designs a structure, he or she must consider manydifferent forms of uncertainty. There are three major types of uncertainty thataffect a structural design:

n There is always substantial uncertainty in predicting theloads a structure might experience at some time in thefuture. Wind, snow, and earthquake loads are highlyunpredictable. The engineer can never be certain of

the maximum number of people that might occupy anapartment building or the weight of the heaviest truckthat might cross a bridge. Truck weights are regulatedby law in the United States, but illegally heavy trucksoccasionally do drive our highways, and it only takesone of them to collapse a bridge. You can post a 20-tonLoad Limit sign on a bridge, but that doesnt mean thedriver of a 30-ton truck wont try to cross it anyway.


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n The strengths of the materials that are used to build actual bridges are also uncertain. Manufacturers of con-struction materials generally pay careful attention to the quality of their products; nonetheless, it is alwayspossible for a batch of substandard steel or concrete to be used in a structure. Even the most conscientiousconstruction contractors occasionally make mistakes on a project, and some construction errors can reduce theability of a structure to carry load.

n The mathematical models we use for structural analysis and design are never 100% accurate. We have alreadyseen this in our discussion of structural modelsactual trusses do not have perfectly straight members orfrictionless pinned connections. Yet we must make these sorts of simplifying assumptions, or the truss analysissimply cannot be performed.

The engineer accounts for all forms of uncertainty by making the structure somewhat stronger than it reallyneeds to beby using a factor of safety in all analysis and design calculations. In general, when it is used in theanalysis of an existing structure, the factor of safety is a dened as

In a truss, the actualforce in a member is called the internal member force, and the force at which failure occursis called the strength. Thus we can rewrite the denition of the factor of safety as

For example, if a structural member has an internal force of 5000 pounds and a strength of 7500 pounds, then itsfactor of safety, FS, is

If the factor of safety is less than 1, then the member or structure is clearly unsafe and will probably fail. If thefactor of safety is 1 or only slightly greater than 1, then the member or structure is nominally safe but has very littlemargin for errorfor variability in loads, unanticipated low member strengths, or inaccurate analysis results. Moststructural design codes specify a factor of safety of 1.6 or larger (sometimes considerably larger) for structural

members and connections.

In Learning Activity #5, well see how the factor of safety is applied in the design process.

On an Actual Bridge Project

Load and Resistance Factor Design

The factor of safety has been used in structural engineering for over a century. In recent years, however, a newdesign philosophy called load and resistance factor design (LRFD) has become increasing popular. LRFD isbased on the idea that the largest loading a structural member experiences in its lifetime must be less than thesmallest possible strength of that member. In an LRFD-based design, the engineer estimates this largestloading by adjusting the loads used in the structural analysis. All loads are multiplied by a code-specied load

factora number that is always greater than 1. The actual magnitude of the load factor depends on howuncertain the loads are. The self-weight of a structure can be predicted accurately, so it has a relatively lowload factor (normally 1.2 to 1.4). Wind, trafc, and earthquake loads are much more unpredictable, so theirload factors are usually much higher. To estimate the smallest possible strength of a member, the engineermultiplies the nominal member strength by a code-specied resistance factora number that is always lessthan 1. The resistance factor accounts for the possibility of understrength materials, fabrication errors, andother uncertainties that may cause a member to be weaker than the engineer intended. Ultimately loadfactors and resistance factors serve the same function as the factor of safetythey ensure the safety of astructure by providing a margin for error. Many experts view the LRFD as a superior design philosophy,because it more accurately represents the sources of uncertainty in structural design.


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The Problem

The Need

One year after the completion of the new Grant Road Bridge, the Hauptville Town Engineer inspects thestructure and nds that it is performing well. Though the bridge has been carrying a lot of trafc, its structuralmembers show no signs of distress or deterioration. Nonetheless, the Town Engineer is still somewhat con-cerned about the bridge. Because of a major construction project nearby, many heavily loaded dump truckshave been using Grant Road recently. What if one of these trucks is heavier than the legal weight limit? Howmuch of an overload would cause the structure to collapse? The Town Engineer decides to perform a completestructural evaluation to determine the overall level of safety of the Grant Road Bridge. He begins by hiringUniversal Structural Materials Assessment, Inc. to test the strength of the structural members used in thebridge. (We did this part of the structural evaluation in Learning Activity #2.) Once the Engineer has receivedthe test results from Universal, he is ready to begin his analysis.

Your Job

You are the Town Engineer of Hauptville. Your job is to analyze the Grant Road Bridge and evaluate itsoverall level of safety. Specically, you must calculate the factor of safety for every member in one of the maintrusses, then determine the overall safety factor for the structure.

As the Town Engineer, you have the professional responsibility to protect the health and safety of thepeople who use this bridge. You fulll this responsibility by performing the structural evaluation consci-entiouslyby using good judgment, by performing calculations carefully and accurately, and by asking a col-league to check your work.

The Solution

The Plan

Our plan to conduct the structural analysis and evaluation of the Grant Road Bridgeconsists of the following tasks:

n Create the structural model.

n Check the structural model for static determinacy and stability.

n Calculate the reactions.

n Calculate the internal member forces.

n Determine the strengths of the members

n Calculate the factor of safety for every member in the structure

n Evaluate the safety of the structure.

n Check our assumptions.

The Learning Activity


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Create the Structural Model

To model the Grant Road Bridge, we must dene (1) the geometry of the structure, (2) the loads, and (3) thesupports and reactions. We begin by idealizing the three-dimensional bridge structure as a pair of two-dimensionalPratt trusses. Since these two trusses are identical, we only need to analyze one of them. The geometry of the trussis shown below. The dimensions indicate the locations of the centerlines of the members. Joints are identied withlettersthe same letter designations that were used on the bridge plans provided in Learning Activity #1. To facili-tate the analysis, we will assume that the truss members are perfectly straight, the joints are frictionless pins, andthe loads are applied only at the joints. We will also assume that the weight of the truss itself is zero.

QQ1Why did we assume that the weight of the truss is zero?

Obviously, the actual weight of the truss is not zero. Why did we

make this assumption, when we know it is not true? How do you

think it will affect the accuracy of our structural analysis?

We will assume that the top chord loading is more severe,

then check the bottom-chord loading later.

When we load-tested the Grant Road Bridge in LearningActivity #1, we applied the load in two different wayswith astack of books placed on the top chord and with a bucket ofsand suspended from the oor beams. Before we can denethe loads for our structural model, we need to decide whichof these two loading congurations to use. As a generalrule, a structural evaluation should be based on the mostsevere loading conditionthe one that produces the highestmember forces. If the analysis shows that the truss is safefor the most severe loading, then the structure will certainlybe safe for less severe ones. Unfortunately, in this case, it isnot immediately obvious which of the two loading con-

gurations is more severe. The best we can do is to make anassumption and check it later. For now, we will assume thatthe top-chord loading, shown here, is more severe.

Having decided on the location of the load, we mustnow determine its magnitude. In Learning Activity #2, weapplied the equation W=mgto determine that the weight of a5-kilogram mass is 49.05 newtons. When we placed the stack of books onto the top chord of the truss, the weight ofthe stack was supported on joints J, J, K, K, L, and L. We can reasonably assume that the weight of the books isdistributed equally to these six joints. Therefore, the downward force applied to each joint is


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Check Static Determinacy and Stability

Before we can use the equations of equilibrium to analyze this truss, we must rst verify that it is staticallydeterminate and stable. The mathematical condition for static determinacy and stability is

wherejis the number of jointsand mis the number of members. Our truss from the Grant Road Bridge has12 joints and 21 members. Substituting these numbers into the equation above, we nd that 2jand m+3are bothequal to24, so the mathematical condition for static determinacy and stability is satised. Furthermore, we notethat the truss is composed entirely of interconnected triangles, which conrms our conclusion that the struc-ture is stable.

Calculate Reactions

On the free body diagram above, the forces RA and RGare the unknown reactions at Joints A and G.We know that the truss is in equilibrium; therefore, the sum of all forces acting on the structure must be zero.Since all of the forcesloads and reactionsare acting in the y-direction, only one of our two equilibriumequations is relevant to the calculation of reactions:

Since there are two main trusses, three 8.175Nloads will be applied to each truss. Since all of the loadsare downward, and the bridge is supported only at its ends, we will add upward reactions RA and RGatjoints A and G.

A complete free body diagram of the truss looks like this:

A

B C D E F

G

I J K L M

8.175N 8.175N 8.175N

RA RG

6 @ 10 cm = 60 cm

12.5 cm

x

y

A

B C D E F

G

I J K L M

8.175N 8.175N 8.175N

RA RG

6 @ 10 cm = 60 cm

12.5 cm

x

y

x

y


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Since the structure, the loads, and the reactions are all symmetrical about the centerline of the truss, the tworeactions RA and RGmust be equal. (The centerline of the truss is a vertical line passing through Member DK).Substituting RA = RGinto the equilibrium equation above, we get

And since RA = RG, then

Calculate Internal Member Forces

We will use the Method of Joints to calculate the internal force in each member of the truss. To apply thismethod, we will isolate a joint from of the structure, cutting through the attached members and exposing theirinternal member forces. We will draw a free body diagram of the joint, then use the equations of equilibrium to

determine the unknown member forces. We will repeat the process for successive joints, until we have calculatedall of the internal member forces in the structure.

Joint AWell start by isolating Joint A and drawing a free body diagram of it. The free body

diagram must show allforces acting on the joint. Thus the reaction RA is shown, along withits known magnitude of 12.26N. The member forces FAIand FABare also included on thediagram. Because we do not know the magnitudes or directions of these forces, we simplyshow them in variable form, and we assume their directions to be in tension. To indicate thata member force is in tension, we draw the force vector pointing away from the joint, alongthe centerline of the member.

Before we can write the equilibrium equations for this joint, we need to gure out whatthe angle is. Actually, we dont really need to know the angle itself; rather, we only reallyneed to know the sine and cosine of the anglesin and cos. We can determine the sineand cosine directly from the geometry of the truss. Note that Members AB, AI, and BI form aright triangle, with Member AI as the hypotenuse. We can apply the Pythagorean Theorem tocalculate the length, LAI, as follows:

Now we can apply the basic denitions of the sine and cosine to nd sin and cos:


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We are nally ready to write the equilibrium equations for Joint A. We will start with the equation for thesum of forces in the y-direction. Assuming that the upward direction is positive,

To write this equation, we had to represent the force FAIin terms of its x-componentand y-component. The y-component is FAIsin, and its direction is upward, so it ispositive in the equilibrium equation. The x-component is FAIcos; however, this com-ponent is not included in the Fyequilibrium equation, because it does not act in they-direction.

We can now substitute the known value of sin into the equilibrium equation, and solve for the unknownforce FAI.

Because the answer is negative, our initial assumption about the direction of FAIwas incorrect. We assumedthat FAIis in tension. The negative member force indicates that it is in compression. Thus our nal answer is

Now we can write the equilibrium equation for forces in the x-direction. Assuming that the positive direc-tion is to the right,

We know cos, and we have just solved for FAI. We can substitute these values into the equilibrium equationand solve for FAB. But be careful! When you substitute FAI, dont forget the minus sign.

Because the answer is positive, our assumption about the direction of FABwas correct.The nal answer is


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QQ2Why did we choose to do Joint A first?

Is there a reason why Joint A was a good place to start this analysis?

What would have happened if we had started with a different joint?

QQ3Why did we choose to solve the y-direction equilibrium equation first?

Is there a reason why it was a good idea to solve Fy= 0beforesolving Fx= 0?

Joint BAt this point, we should analyze Joint B. It has only three connected members, and we already know the inter-

nal force in one of the three (Member AB). Thus there are only two unknown forces, and we will be able to solvefor them with the two available equilibrium equations.

Again we draw a free body diagram of the joint, with all member forces assumed tobe in tensionpointing away from the joint. The known magnitude of the force FABisincluded on the diagram. The equilibrium equation for forces in the x-direction is

The equilibrium equation for forces in the y-direction produces an interesting result:

It should come as no surprise that this member has zero internal force. When you load-tested the Grant RoadBridge in Learning Activity #1, you should have noticed that this memberthe hip verticalwas slack. It appearedto have no internal force at all. Now we have veried our observation using the Method of Joints!

Joint IIt takes some careful thought to recognize that Joint I should be the next joint we

analyze. As the free body diagram indicates, this joint has four connected membersand, therefore, it also has four internal member forces. Note, however, that we havealready calculated two of theseFAIand FBI. Thus there are only two unknown forces,which we can calculate with our two equilibrium equations.


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QQ4Why Joint C?

Why was Joint C the best joint to analyze at this point in the

solution process?

Note that all four of the force vectors are pointing away from the joint, even though we already know thatone of them, FAI, is in compression. The negative magnitude of FAIensures that it is mathematically representedas a compression force.

It is important to note that both angles labeled as on the free body diagram are exactly the same as theangle on the diagram of Joint A. (If you cant see that these angles are all equal, prove it to yourself by drawingthe corresponding right triangles, just as we did for Joint A.) Thus the values we calculated for sin and cos forJoint A are still valid here.

If we begin with the equilibrium equation in the y-direction, we will be able to solve for FCIdirectly:

Now we can use the second equilibrium equation to solve for FIJ:

Joint CNext we will calculate the unknown member forces at Joint C.

Based on the free body diagram of the joint, we can write the two equi-librium equations and solve for the two unknown member forces as follows:


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Joint JThe free body diagram of Joint J is shown at right. Note that the 8.175N

load at Joint J mustbe included on the diagram. (Failure to put loads on thefree body diagram is one of the most common errors in truss analysis.) Wecan write the two equilibrium equations and solve for the two unknownmember forces as follows:


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Summary of Structural Analysis ResultsAt this point, we have only analyzed half of the truss. However, if we take advantage of symmetry, we can

determine the internal forces in all remaining members without doing any further calculations. When wedetermined the reactions RA and RG, we noted that these two forces must be equal because the structure, itsloads, and its reactions are all symmetrical about the centerline of the truss. The same principle holds true forinternal member forces. Because the structure, loads, and reactions are all symmetrical, the member forcesmust also be symmetrical about the centerline. Members that are mirror images of each other have equalinternal forces. FGMand FAImust be equal; FLMand FIJmust be equal; FDJand FDLmust be equal, and so forth.

So were done! The results of the analysis are summarized in the table below.

Determine the Strengths of the Members

Now that we have calculated the force in each member, we must determine the corresponding strength ofeach member. To do this, we will use the graphs we developed in Learning Activity #2. Well start with thebarsthe bottom chords, diagonals, and hip verticals. The table above tells us what we have already observedin our Grant Road Bridge modelthat all of the bars are in tension (except the hip verticals, which have zero

QQ5Can you apply the Method of Joints to calculate a member force?

Which joint should you analyze to determine the member force FDK?

Solve the appropriate equilibrium equations to show that FDK=8.175N

(compression).

Members Force Members Force

AB, FG 9.81 N (tension) BI, FM 0 NBC, EF 9.81 N (tension) CI, EM 15.70 N (tension)CD, DE 19.62 N (tension) CJ, EL 12.26 N (compression)IJ, LM 19.62 N compression) DJ, DL 5.23 N (tension)JK, KL 22.89 N (compression) DK 8.175 N (compression)AI, GM 15.70 N (compression)


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internal force). Thus we need to determine the tensile strength of the bars. All of the bars used in the Grant RoadBridge are 4mm wide. Using the tensile strength vs. member width graph we developed in Learning Activity #2, wend the tensile strength of a 4mm bar to be 26 newtons, as shown below.

0

10

20

30

40

60

0 2 4 6 8 12 14 16 18

Length (cm)

CompressiveStrength(newtons)

10mm x 10mm tube

6mm x 10mm tube

0

10

20

30

40

60

0 2 4 6 8 12 14 16 18

Length (cm)


10mm x 10mm tube

6mm x 10mm tube50

10

0.0

10.0

20.0

30.0

40 .0

50.0

60.0

0 1 2 3 4 5 6 7 8 9

TensileStrength(newtons

Trend Line

26

0.0

10.0

20.0

30.0

40 .0

50.0

60.0

0 1 2 3 4 5 6 7 8 9

M e m b e r W i d t h ( m m )

Trend Line

26

Determining the tensile strength of a 4mm bar

Determining the compressive strength of a 10mm x 10mm tube that is10 centimeters long

Note, however, that all of the bottom chords, diagonals, and hip verticals are actually doubled4mm bars. Thusthe tensile strength of these members is exactly twice that of a single 4mm bar, or 52 newtons.

Our structural analysis shows that all of the tubesthe top chords, the end posts, and the interior verticalsarein compression. Thus we must determine the compressive strength of these members, using the strength vs. lengthgraph we developed in Learning Activity #2. The top chord members are all 10mm x 10mm tubes, and each has a

length of 10 centimeters. The strength of these members is approximately 50 newtons, as indicated below.


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QQ6Can you determine the strength of a member?

What is the compressive strength of the vertical tube members (CJ,

DK, and EL) and the end posts (AI and GM)?

QQ7Can you calculate the factor of safety for a member?

Calculate the factor of safety for all remaining members in the truss,

and add them to the summary table below (along with the member

strengths not already recorded in the table).

Calculate the Factor of Safety

Once we know the strength of a member and the internal force it is actually experiencing, we can calculateits factor of safety. For the bottom chord member CD, the factor of safety is:

For the top chord member JK, the factor of safety is

Members Force Strength FS

AB, FG 9.81 N (tension) 52

BC, EF 9.81 N (tension) 52

CD, DE 19.62 N (tension) 52 2.7

IJ, LM 19.62 N (compression) 50

JK, KL 22.89 N (compression) 50 2.2

AI, GM 15.70 N (compression)

BI, FM 0 N -- 52 --CI, EM 15.70 N (tension) 52

CJ, EL 12.26 N (compression)

DJ, DL 5.23 N (tension) 52

DK 8.175 N (compression)


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QQ8Why are some truss members stronger than they need to be?

For example, Member CI is a doubled 4mm bar with a safety factor of

3.3. Had the structural engineer chosen to use a doubled 3mm bar for

this member, the safety factor would still be 2.5. The member would

use less material; and because its safety factor would still be greater

than 2.2, the overall safety of the structure would not be adversely

affected. Why did the structural engineer choose not to use a smaller

member size?

Evaluate the Structure

As the Town Engineer of Hauptville, you have nished what you set out to doa complete structural evaluationof the main trusses of the Grant Road Bridge. Yet the results of these calculations are just numbers. They are oflittle use, until you study them, think critically about them, and draw meaningful conclusions from them.

Once you have completed the summary table on the previous page, you should be able to make the followingobservations:

n Members JK and KL have a factor of safety of 2.2the smallest of any member in the truss. Since the failure ofMember JK or Member KL would cause the entire structure to collapse, we can say that the factor of safety ofthe entire structure is 2.2.

n Since 2.2 is obviously larger than 1, our analysis tells us that the structure will not collapse when the 5 kg mass isplaced on the top chord. Because the factor of safety is considerably larger than 1, we can have a high degree ofcondence that the structure will not fail, even if we made some minor errors in construction or if the actualload is signicantly larger than 5 kg.

n Theoretically, the bridge would collapse if the mass of the stack of books at mid-span were increased to(5 kg)(2.2)=11.0 kg.

n Many members of the truss have safety factors that are substantially larger than 2.2. These members are actuallymuch stronger than they need to be.

It is important to note that this structural evaluation is valid only for one particular loading. If we change eitherthe magnitude or the position of the load, the member forces and factors of safety will also change.


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On an Actual Bridge Project

Deections are important too.

We evaluated the Grant Road Bridge by checking that each

member in the structure can carry load safely. On an actual

bridge project, the engineer would also check to ensure that the deections are accept-ably small. A deection is the distance a structure moves when it is loaded. A bridge

always deects when it is loaded. On a well-designed bridge, these deections are so

small that they are imperceptible to drivers and pedestrians crossing the span. If a bridge

is too exible, then drivers and pedestrians will feel its movement and will perceive the

structure to be unsafeeven if its strength entirely adequate. The engineer is responsible

for ensuring not only that the bridge issafe, but also that the publicperceives it to besafe.

Thus the engineer carefully computes the structures deections under various loading

conditions and ensures that these computed deections comply with the appropriate

design codes.

Bottom-Chord Loading

The analysis above was based on a number ofassumptions. Perhaps the most important of these wasour assumption that placing the load on the top chordof the truss is more severe than suspending the loadfrom the oor beams. We now have the analytical toolsto check this assumption. Since the oor beams areattached to the bottom chords, loading the oor beams

is essentially the same as loading the bottom chordjoints of the truss. Thus our revised structural modelshould have the three 8.175 N loads applied at thebottom chord, as shown below:

AB C D E F

G

I J K L M

8.175N 8.175N 8.175N RA RG

AB C D E F

G

I J K L M

8.175N 8.175N 8.175N RA RG

Members Force Members Force

AB, FG 9.81 N (tension) BI, FM 0 NBC, EF 9.81 N (tension) CI, EM 15.70 N (tension)CD, DE 19.62 N (tension) CJ, EL 4.09 N (compression)IJ, LM 19.62 N compression) DJ, DL 5.23 N (tension)JK, KL 22.89 N (compression) DK 0 N (compression)AI, GM 15.70 N (compression)

Is bottom-chord loading more or less

severe than top-chord loading?

If we repeat the structural analysis using this new loading condition, we get the following results:


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QQ9Can you apply the Method of Joints to analyze a truss?

Use the Method of Joints to analyze the Pratt truss with bottom-chord

loading. Prove that the results in the table above are correct.

Note that moving the loads from the top chord to the bottom chord caused the internal forces to change onlyinMembers CJ, EL, and DK. In all three cases, the forces got smaller. The forces in the most heavily loaded mem-bersthe top and bottom chords and the end postsremained unchanged; thus the overall factor of safety of thestructure remains unchanged. We can conclude that using the top-chord loading in Learning Activity #1 (and in ourinitial structural analysis) was entirely appropriate. The bottom-chord loading is more realistic, because real bridgescarry trafc loads on their oor beams; however, the top-chord loading is much easier to do and produces nearlyidentical structural analysis results.

QQ1 0Can members with zero force be removed?

In the analysis above, Members BI, FM, and DK have zero internal

force. It would seem that the truss does not need these members to

carry load, and we might simply remove them from the structure to

save material. Do you think these zero-force members can safely be

removed from the truss?

QQ1 1Can you analyze a different truss?

Select any truss from the Gallery of Structural Analysis Results

(Appendix B), and calculate the internal forces in all of its members.

Conclusion

In this learning activity, we applied concepts from geometry, trigonometry, algebra, and physics to calculate theinternal forces in every member of a truss. In doing so, you saw how math and science are applied to solve animportant, real-world problem. You also saw how data from laboratory experimentsthe strength tests we did inLearning Activity #2can be integrated into the solution of an engineering problem. Most important, you learnedto use an important analysis tool called the Method of Joints. This technique isnt easy! It requires you to con-struct and solve equilibrium equationslots of themwhile paying careful attention to the magnitudes and direc-tions of forces. Wouldnt it be great if we could use the computer to do this work for us? Well, we can. In LearningActivity #4, we will use the West Point Bridge Designer software to analyze and evaluate a truss bridge.


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Answers to the Questions

1) Why did we assume that the weight of the truss is zero? We couldinclude the self-weight of the bridge inour structural analysis, but doing so wouldgreatlycomplicate the analysis. In Learning Activity #1, we foundthat the mass of our Grant Road Bridge model is about 55 grams. This mass is very small in comparison with the5 kilogram mass the bridge is designed to carry. Including the self-weight in our analysis would change ourcalculated member forces by only about one percent. For the sake of simplicity, we can ignore self-weight, andrecognize that this assumption has a very small effect on the accuracy of the structural analysis.

2) Why did we choose to do Joint A first? At any given joint, we can write two equilibrium equationsFx=0and Fy=0. With only two available equations, we can solve for only two unknown member forces at each joint.At the start of the solution process, Joint A has just two unknown member forcesFABand FAI. Joint G is theonly other joint in the entire truss with only two unknown forces. All of the others have three or more. Thusthe solution process should start at either Joint A or Joint G.

Note that the reaction force RAis also applied at Joint A. Had this force also been unknown, it would havebeen impossible to solve the equilibrium equations at this jointthere would have been three unknowns andonly two equations. Thats why it is generally necessary to solve for reactions before we use the Method ofJoints to calculate member forces.

3) At Joint A, why did we choose to solve the y-direction equilibrium equation first? At Joint A, the x-directionequilibrium equation has two unknown member forces, FABand FAI. Had we written this equation rst, we

would have been unable to solve for either of the unknowns immediately. The y-direction equation includesonly one unknown force, FAI. By solving Fy=0rst, we were able to solve for FAIdirectly. Then, when we wroteFx=0, we could substitute the known value of FAI and solve for FAB. At any given joint, we can often (but notalways) avoid the chore of solving two equations simultaneously by identifying an equilibrium equation that hasonly one unknown member forceand solving it rst.

4) After solving for the unknown member forces at Joints A, B, and I, why did we choose Joint C? At Joints A, B,and I we calculated the member forces FAB, FAI, FBC, FBI, FCI, and FIJ. At this point in the solution process, wealready knew two of the four member forces at Joint CFBCand FCI. Only FCDand FCJwere unknown. So weselected Joint C because it had only two unknown member forces, which could be solved with our two equa-tions of equilibrium. As a general rule, when using the Method of Joints to analyze a truss, always select a jointwith only two unknown member forces as the next step in the analysis.

5) Can you apply the Method of Joints to calculate the force in Member DK? The

best joint to analyze in order to determine the member force FDKis Joint K. The freebody diagram of this joint is shown at right. From this diagram, we can calculate FDKdirectly from the y-direction equilibrium equation

0Fy=


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6) Can you determine the strength of the verticals and end posts? To determine the compressive strength of theverticals and end posts, we will use the strength vs. length graph we developed in Learning Activity #2. The verticalsare 10mm x 6mm tubes, and each has a length of 12.5 centimeters. Thus the compressive strength of the verticaltubes is 43 newtons, as indicated below:

Determining the compressive strength of a 6mm x 10mm tube that is 12.5cm long

The end posts are 10mm x 10mm tubes. Since they are diagonal members, we must calculate their length usingthe Pythagorean Theorem:

As the graph below indicates, a 10mm x 10mm tube with a length of 16 centimeters has a compressive strength

of 42 newtons.

0

10

20

30

40

50

60

0 2 4 6 8 10 14 16 18

Length (cm)

CompressiveStrength(newton

s)

10mm x 10mm tube

6mm x 10mm tube

0

10

20

30

40

50

60

0 2 4 6 8 10 14 16 18

Length (cm)

CompressiveStrength(newton

s)

10mm x 10mm tube

6mm x 10mm tube

43

12.5

0

10

20

30

40

50

60

0 2 4 6 8 10 12 14 18

Length (cm)


10mm x 10mm tube

6mm x 10mm tube

42

0

10

20

30

40

50

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0 2 4 6 8 10 12 14 18

Length (cm)


10mm x 10mm tube

6mm x 10mm tube

42

16

Determining the compressive strength of a 10mm x 10mm tube that is 16cm long


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7) Can you calculate the factor of safety for all of the remaining truss members? Once the internal memberforce and strength are known for each member in the truss, the corresponding factor of safety can be calculatedusing the equation

The results are summarized in the table below.

Members Force Strength FS

AB, FG 9.81 N (tension) 52 5.3

BC, EF 9.81 N (tension) 52 5.3

CD, DE 19.62 N (tension) 52 2.7

IJ, LM 19.62 N (compression) 50 2.5

JK, KL 22.89 N (compression) 50 2.2

AI, GM 15.70 N (compression) 42 2.7

BI, FM 0 N -- 52 --

CI, EM 15.70 N (tension) 52 3.3CJ, EL 12.26 N (compression) 43 3.5

DJ, DL 5.23 N (tension) 52 9.9

DK 8.175 N (compression) 43 5.3

8) Why are some truss members stronger than they need to be? For example, why did the structural engineernot use a doubled 3mm bar for Member CI, rather than the doubled 4mm bar she specied in the design?Making this change would not adversely affect the overall safety of the structure; yet reducing the member sizewould clearly use less material. And using less material ought to lower the cost of the structure, right?

Member CI and a number of other members in the Grant Road Bridge are too strong because the struc-tural engineer chose to use only a limited number of different member sizes in her design. In this case, sheused only threethe two tubes and the doubled 4mm bar. She chose the 4mm bar so that the tension mem-bers with the largest internal forceCD and DEwould have a factor of safety greater than 2. Then she simplyspecied the same 4mm bar for all of the other tension members, knowing that this size would be more thanadequate for members whose internal forces were lower.

To understand why the engineer chose to use a limited number of member sizes, just think about your ownexperience building the Grant Road Bridge. Suppose the engineer had designed every member with a safetyfactor of exactly 2. Most likely, each main truss would have required ve different bar sizes and ve differenttube sizes. With so many different sizes, it would have taken you much longer to measure, cut out, and assem-ble the members. The connections would also have been much more complicated, and you would have beenmore likely to make construction errorsto put a 3mm bar where a 4mm bar is supposed to be used, for

example. You would have saved some material, because every member would only be as strong at it absolutelyneeds to be. But this small reduction in material cost probably would not have been worth all of the extra work.

The same is true for the construction of a real structurethere can be substantial cost saving in using a fewstandard member sizes, because doing so can greatly simplify the fabrication and construction processes.


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9) Can you apply the Method of Joints to analyze a truss? To analyze the truss with bottom-chord loading, followexactly the same procedure as you did for the truss with top-chord loading. You will nd that the calculation ofreactions, and the analyses of Joints A, B, and I are exactlythe same as for the top-chord loading. Starting at Joint C,however, the two solutions differ. The truss with bottom-chord loading has an 8.175N load at Joint C, as shown inthe free body diagram. As a result, the x-direction equilibrium equation remains the same, but the y-directionequilibrium equation changes:

Next we analyze Joint J. In this case, the absence of anexternal load at the joint causes the y-direction equilibriumequation to change.

Finally we analyze Joint K to determine FDK

.


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10) Can members with zero force be removed? Members BI and FM could be removed from our model of theGrant Road Bridge with no adverse consequences; however, they could not be safely removed from the actualhighway bridge. In the actual bridge, Members BI and FM serve an important structural functionthey supportthe oor beams attached to the truss at Joints B and F. These oor beams help to support the bridge deck andtransmit vehicular loads from the deck to the main trusses. Thus, in the actual bridge, Members BI and FM arein tension and are essential to the structures load-carrying ability.

Member DK could not be safely removed from either the model or the actual bridge. Without Member DK,we would have just one continuous member from Joint J to Joint Lthere would be no reason for a joint at K.

This new memberlets call it Member JLwould be twice as long as Member JK or Member KL. As a result,Member JL would be much weaker in compression than JK or KL. (Recall from Learning Activity #2 that com-pression strength decreases substantially with increasing length.) To keep Member JL from buckling, a con-siderably larger tube would be required. Thus, even though Member DK has no internal force, it effectivelystrengthens the structure by dividing Member JL into two shorter, stronger compression members.

It is also worth noting that, in an actual bridge, the internal force in Member DK would not be zero. Itwould actually carry part of the self-weight of Members JK and KL, resulting in a small compressive internalforce. In our analysis, the internal force in Member DK is zero only because we assumed the self-weight to bezero.

11) Can you analyze a more complex truss? The Gallery of Structural Analysis Results (Appendix B) providesthe calculated internal member forces for a wide variety of common truss congurations.

Analyze & Evaluate a Truss

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